You don't need locally compact or Hausdorff. Let $f, g \in C_0(X)$. Given any $\epsilon > 0$, we want to show that $C = \{ x \in X: |f(x) + g(x)| \geq \epsilon\}$ is compact. Let $$A = \{ x \in X : |f(x)| \geq \epsilon/3 \}$$ $$B = \{x \in X : |g(x)| \geq \epsilon/3\}.$$
These are both compact, and hence so is $A \cup B$.
The triangle inequality shows that $C$ is contained in $A \cup B$, making it compact as well. This establishes that $f+g \in C_0(X)$. A similar argument shows that $fg \in C_0(X)$.
A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.
Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.
Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.
Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.
Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.
Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.
Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.
Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.
Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.
Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.
Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.
Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.
Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.
Best Answer
Let $e\in C_0(X)$ be an identity and fix any $x_0\in X$. We will show that $e(x_0)=1$, whence $e=1$.
Since $X$ is locally compact, there is a neigbhourhood $K$ of $x_0$ which is compact, and so it is closed because $X$ is Hausdorff. Moreover, since $X$ is Hausdorff, $K$ is compact Hausdorff, and thus it is normal. Let $F$ be the boundary of $K$. Since $K$ is closed, $F\subseteq K$.
Now, $F$ and $\{x_0\}$ are closed in $K$, so by Urysohn's lemma, there is a continuous real-valued $f\colon K\to \mathbf R$ such that $f(x_0)=1$ and $f$ vanishes on $F$. But then $f$ can be extended to $\bar f\colon X\to \mathbf R$ which vanishes outside of $K$.
Finally, it is easy to see that $\bar f$ is continuous, and it is compactly supported, so it clearly is in $C_0(X)$. Since $\bar f\cdot e(x_0)=1=\bar f(x_0)$, it follows that $e(x_0)=1$.